Answer
Verified
327.6k+ views
Hint :The $ N{a_2}C{O_3} $ is sodium carbonate, which is being converted in sodium bicarbonate $ \left( {NaHC{O_3}} \right). $ $ A{l_2}{\left( {S{O_4}} \right)_3} $ is aluminum sulphate, and is being converted to aluminum hydroxide $ \left( {Al{{\left( {OH} \right)}_3}} \right). $ $ N{a_2}{S_2}{O_3} $ is sodium thiosulphate is converted to tetrathionate anion $ \left( {{S_4}O_6^{2 - }} \right). $
Complete Step By Step Answer:
We have:
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
For $ N{a_2}C{O_3} \to NaHC{O_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ N{a_2}C{O_3}, $ carbonate anion $ \left( {CO_3^{2 - }} \right) $ has $ - 2 $ charge. This means that the total positive charge contributed by the cations $ \left( {2N{a^ + }} \right) $ is $ + 2. $ In $ NaHC{O_3}, $ bicarbonate anion has $ - 1 $ charge, and hence total positive charge contributed by cation $ \left( {N{a^ + }} \right) $ is $ + 1. $
Therefore, n-factor value $ = 2 - 1 = 1 $
(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
For $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ A{l_2}{\left( {S{O_4}} \right)_3}, $ sulphate anion $ \left( {SO_4^{2 - }} \right) $ has $ - 2 $ charge. The total negative charge will therefore be $ - 2 \times 3 = 6 $ as there are three sulphate ions. This means that the total positive charge contributed by the cations $ \left( {2A{l^{3 + }}} \right) $ is $ + 6. $ In $ Al{\left( {OH} \right)_3}, $ hydroxide ion has $ - 1 $ charge, and hence three hydroxide ions contribute $ - 3 $ charge. The total positive charge contributed by cation $ \left( {A{l^{3 + }}} \right) $ is $ + 3. $
Therefore, n-factor value $ = 6 - 3 = 3 $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
For $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $ the n-factor is the product of change in oxidation state and number of atoms undergoing change. Since the change undergone is as follows: $ {S_2}O_3^{2 - } \to {S_4}O_6^{2 - } $ we need to calculate the oxidation state of sulphur atom in each of these. The oxidation states of sulphur in $ {S_2}O_3^{2 - } $ and $ {S_4}O_6^{2 - } $ are $ 2 $ and $ 2.5 $ respectively and hence the change in oxidation state is $ 2.5 - 2 = 0.5. $ The number of sulphur atoms undergoing change are $ 4 - 2 = 2. $
Therefore, n-factor value $ = 0.5 \times 2 = 1 $
Hence, the required answer is (B) $ 1,3,1. $
Note :
The n-factor can be calculated for different compounds according to their characteristics: whether they are an acid, base, salt or if the reaction undergone by the compound is a redox reaction. For acids, the n-factor is represented in the form of basicity, while for bases, the n-factor is represented in the form of acidity. For salts, the number of positive or negative ions are counted. For redox reactions, the change in oxidation number or change in their reduction number in both sides of the reaction is considered.
Complete Step By Step Answer:
We have:
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
For $ N{a_2}C{O_3} \to NaHC{O_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ N{a_2}C{O_3}, $ carbonate anion $ \left( {CO_3^{2 - }} \right) $ has $ - 2 $ charge. This means that the total positive charge contributed by the cations $ \left( {2N{a^ + }} \right) $ is $ + 2. $ In $ NaHC{O_3}, $ bicarbonate anion has $ - 1 $ charge, and hence total positive charge contributed by cation $ \left( {N{a^ + }} \right) $ is $ + 1. $
Therefore, n-factor value $ = 2 - 1 = 1 $
(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
For $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ A{l_2}{\left( {S{O_4}} \right)_3}, $ sulphate anion $ \left( {SO_4^{2 - }} \right) $ has $ - 2 $ charge. The total negative charge will therefore be $ - 2 \times 3 = 6 $ as there are three sulphate ions. This means that the total positive charge contributed by the cations $ \left( {2A{l^{3 + }}} \right) $ is $ + 6. $ In $ Al{\left( {OH} \right)_3}, $ hydroxide ion has $ - 1 $ charge, and hence three hydroxide ions contribute $ - 3 $ charge. The total positive charge contributed by cation $ \left( {A{l^{3 + }}} \right) $ is $ + 3. $
Therefore, n-factor value $ = 6 - 3 = 3 $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
For $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $ the n-factor is the product of change in oxidation state and number of atoms undergoing change. Since the change undergone is as follows: $ {S_2}O_3^{2 - } \to {S_4}O_6^{2 - } $ we need to calculate the oxidation state of sulphur atom in each of these. The oxidation states of sulphur in $ {S_2}O_3^{2 - } $ and $ {S_4}O_6^{2 - } $ are $ 2 $ and $ 2.5 $ respectively and hence the change in oxidation state is $ 2.5 - 2 = 0.5. $ The number of sulphur atoms undergoing change are $ 4 - 2 = 2. $
Therefore, n-factor value $ = 0.5 \times 2 = 1 $
Hence, the required answer is (B) $ 1,3,1. $
Note :
The n-factor can be calculated for different compounds according to their characteristics: whether they are an acid, base, salt or if the reaction undergone by the compound is a redox reaction. For acids, the n-factor is represented in the form of basicity, while for bases, the n-factor is represented in the form of acidity. For salts, the number of positive or negative ions are counted. For redox reactions, the change in oxidation number or change in their reduction number in both sides of the reaction is considered.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Draw a diagram showing the external features of fish class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE